\(\int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} (-4+5 e^{5 x/4} x)}{16 x^2} \, dx\) [5190]

Optimal result

Integrand size = 45, antiderivative size = 28 \[ \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx=4+\frac {e^{e^{5 x/4}}}{4 x}+x+x \left (-x+x^5\right ) \]

[Out]

4+1/4*exp(exp(5/4*x))/x+x*(x^5-x)+x

Rubi [F]

\[ \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx=\int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx \]

[In]

Int[(16*x^2 - 32*x^3 + 96*x^7 + E^E^((5*x)/4)*(-4 + 5*E^((5*x)/4)*x))/(16*x^2),x]

[Out]

x - x^2 + x^6 - Defer[Int][E^E^((5*x)/4)/x^2, x]/4 + (5*Defer[Int][E^(E^((5*x)/4) + (5*x)/4)/x, x])/16

Rubi steps \begin{align*} \text {integral}& = \frac {1}{16} \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{x^2} \, dx \\ & = \frac {1}{16} \int \left (\frac {5 e^{e^{5 x/4}+\frac {5 x}{4}}}{x}+\frac {4 \left (-e^{e^{5 x/4}}+4 x^2-8 x^3+24 x^7\right )}{x^2}\right ) \, dx \\ & = \frac {1}{4} \int \frac {-e^{e^{5 x/4}}+4 x^2-8 x^3+24 x^7}{x^2} \, dx+\frac {5}{16} \int \frac {e^{e^{5 x/4}+\frac {5 x}{4}}}{x} \, dx \\ & = \frac {1}{4} \int \left (-\frac {e^{e^{5 x/4}}}{x^2}+4 \left (1-2 x+6 x^5\right )\right ) \, dx+\frac {5}{16} \int \frac {e^{e^{5 x/4}+\frac {5 x}{4}}}{x} \, dx \\ & = -\left (\frac {1}{4} \int \frac {e^{e^{5 x/4}}}{x^2} \, dx\right )+\frac {5}{16} \int \frac {e^{e^{5 x/4}+\frac {5 x}{4}}}{x} \, dx+\int \left (1-2 x+6 x^5\right ) \, dx \\ & = x-x^2+x^6-\frac {1}{4} \int \frac {e^{e^{5 x/4}}}{x^2} \, dx+\frac {5}{16} \int \frac {e^{e^{5 x/4}+\frac {5 x}{4}}}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx=\frac {e^{e^{5 x/4}}}{4 x}+x-x^2+x^6 \]

[In]

Integrate[(16*x^2 - 32*x^3 + 96*x^7 + E^E^((5*x)/4)*(-4 + 5*E^((5*x)/4)*x))/(16*x^2),x]

[Out]

E^E^((5*x)/4)/(4*x) + x - x^2 + x^6

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
risch	\(x^{6}-x^{2}+x +\frac {{\mathrm e}^{{\mathrm e}^{\frac {5 x}{4}}}}{4 x}\)	\(21\)
norman	\(\frac {x^{2}+x^{7}-x^{3}+\frac {{\mathrm e}^{{\mathrm e}^{\frac {5 x}{4}}}}{4}}{x}\)	\(24\)
parallelrisch	\(\frac {16 x^{7}-16 x^{3}+16 x^{2}+4 \,{\mathrm e}^{{\mathrm e}^{\frac {5 x}{4}}}}{16 x}\)	\(29\)

[In]

int(1/16*((5*x*exp(5/4*x)-4)*exp(exp(5/4*x))+96*x^7-32*x^3+16*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x^6-x^2+x+1/4*exp(exp(5/4*x))/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx=\frac {4 \, x^{7} - 4 \, x^{3} + 4 \, x^{2} + e^{\left (e^{\left (\frac {5}{4} \, x\right )}\right )}}{4 \, x} \]

[In]

integrate(1/16*((5*x*exp(5/4*x)-4)*exp(exp(5/4*x))+96*x^7-32*x^3+16*x^2)/x^2,x, algorithm="fricas")

[Out]

1/4*(4*x^7 - 4*x^3 + 4*x^2 + e^(e^(5/4*x)))/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx=x^{6} - x^{2} + x + \frac {e^{e^{\frac {5 x}{4}}}}{4 x} \]

[In]

integrate(1/16*((5*x*exp(5/4*x)-4)*exp(exp(5/4*x))+96*x**7-32*x**3+16*x**2)/x**2,x)

[Out]

x**6 - x**2 + x + exp(exp(5*x/4))/(4*x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx=x^{6} - x^{2} + x + \frac {e^{\left (e^{\left (\frac {5}{4} \, x\right )}\right )}}{4 \, x} \]

[In]

integrate(1/16*((5*x*exp(5/4*x)-4)*exp(exp(5/4*x))+96*x^7-32*x^3+16*x^2)/x^2,x, algorithm="maxima")

[Out]

x^6 - x^2 + x + 1/4*e^(e^(5/4*x))/x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).

Time = 0.32 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx=\frac {{\left (4 \, x^{7} e^{\left (\frac {5}{4} \, x\right )} - 4 \, x^{3} e^{\left (\frac {5}{4} \, x\right )} + 4 \, x^{2} e^{\left (\frac {5}{4} \, x\right )} + e^{\left (\frac {5}{4} \, x + e^{\left (\frac {5}{4} \, x\right )}\right )}\right )} e^{\left (-\frac {5}{4} \, x\right )}}{4 \, x} \]

[In]

integrate(1/16*((5*x*exp(5/4*x)-4)*exp(exp(5/4*x))+96*x^7-32*x^3+16*x^2)/x^2,x, algorithm="giac")

[Out]

1/4*(4*x^7*e^(5/4*x) - 4*x^3*e^(5/4*x) + 4*x^2*e^(5/4*x) + e^(5/4*x + e^(5/4*x)))*e^(-5/4*x)/x

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {16 x^2-32 x^3+96 x^7+e^{e^{5 x/4}} \left (-4+5 e^{5 x/4} x\right )}{16 x^2} \, dx=x-x^2+x^6+\frac {{\mathrm {e}}^{{\left ({\mathrm {e}}^x\right )}^{5/4}}}{4\,x} \]

[In]

int(((exp(exp((5*x)/4))*(5*x*exp((5*x)/4) - 4))/16 + x^2 - 2*x^3 + 6*x^7)/x^2,x)

[Out]

x - x^2 + x^6 + exp(exp(x)^(5/4))/(4*x)